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Question Number 35363 by Rio Mike last updated on 18/May/18

Q_1 . A quadratic equation x^2 −3x+4=0 has roots α and β .without solving a)write down the values of α^2 +β^2 b) find the quadratic equation with integral coefficients,whose roots are (1/α^2 ) and(1/β^2 ) Q_2 . the first term of a GP is 32 and the sum to infinity is 48. find the common ratio and the 8^(th) term of the progression

$${Q}_{\mathrm{1}} .\:{A}\:{quadratic}\:{equation}\:{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{4}=\mathrm{0} \\ $$$${has}\:{roots}\:\alpha\:{and}\:\beta\:.{without}\:{solving} \\ $$$$\left.{a}\right){write}\:{down}\:{the}\:{values}\:{of}\:\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \\ $$$$\left.{b}\right)\:{find}\:{the}\:{quadratic}\:{equation} \\ $$$${with}\:{integral}\:{coefficients},{whose} \\ $$$${roots}\:{are}\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:{and}\frac{\mathrm{1}}{\beta^{\mathrm{2}} } \\ $$$${Q}_{\mathrm{2}} .\:{the}\:{first}\:{term}\:{of}\:{a}\:{GP}\:{is}\:\mathrm{32}\: \\ $$$${and}\:{the}\:{sum}\:{to}\:{infinity}\:{is}\:\mathrm{48}. \\ $$$${find}\:{the}\:{common}\:{ratio}\:{and}\:{the}\:\mathrm{8}^{{th}} \\ $$$${term}\:{of}\:{the}\:{progression} \\ $$

Answered by Rasheed.Sindhi last updated on 18/May/18

Q_1 α & β are the roots of x^2 −3x+4=0 α+β=((−(−3))/1)=3 , αβ=(4/1)=4 (a) α^2 +β^2 =(α+β)^2 −2αβ =(3)^2 −2(4)=9−8=1 (b) Thwe equation whose roots are (1/α^2 ) &(1/β^2 ) Sum-of-roots= (1/α^2 )+(1/β^2 )=((α^2 +β^2 )/((αβ)^2 )) =(1/((4)^2 ))=1/16 Product-of-roots= (1/α^2 )×(1/β^2 )=(1/((αβ)^2 )) =(1/((4)^2 ))=(1/(16)) The new equation: x^2 −(sum)x+product=0 x^2 −(1/16)x+(1/16)=0 16x^2 −x+1=0

$$\mathrm{Q}_{\mathrm{1}} \\ $$$$\alpha\:\&\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\alpha+\beta=\frac{−\left(−\mathrm{3}\right)}{\mathrm{1}}=\mathrm{3}\:,\:\alpha\beta=\frac{\mathrm{4}}{\mathrm{1}}=\mathrm{4} \\ $$$$\left({a}\right)\:\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{3}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{4}\right)=\mathrm{9}−\mathrm{8}=\mathrm{1} \\ $$$$\left({b}\right)\:\mathrm{Thwe}\:\mathrm{equation}\:\mathrm{whose}\:\mathrm{roots}\:\mathrm{are}\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:\&\frac{\mathrm{1}}{\beta^{\mathrm{2}} } \\ $$$$\mathrm{Sum}-\mathrm{of}-\mathrm{roots}=\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }+\frac{\mathrm{1}}{\beta^{\mathrm{2}} }=\frac{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }{\left(\alpha\beta\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\left(\mathrm{4}\right)^{\mathrm{2}} }=\mathrm{1}/\mathrm{16} \\ $$$$\mathrm{Product}-\mathrm{of}-\mathrm{roots}=\:\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }×\frac{\mathrm{1}}{\beta^{\mathrm{2}} }=\frac{\mathrm{1}}{\left(\alpha\beta\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\left(\mathrm{4}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$\:\mathrm{The}\:\mathrm{new}\:\mathrm{equation}: \\ $$$$\:\:\:\mathrm{x}^{\mathrm{2}} −\left(\mathrm{sum}\right)\mathrm{x}+\mathrm{product}=\mathrm{0} \\ $$$$\:\:\mathrm{x}^{\mathrm{2}} −\left(\mathrm{1}/\mathrm{16}\right)\mathrm{x}+\left(\mathrm{1}/\mathrm{16}\right)=\mathrm{0} \\ $$$$\:\mathrm{16x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}=\mathrm{0} \\ $$

Commented by MJS last updated on 18/May/18

you′re typing faster than me ;−)

$$\left.\mathrm{you}'\mathrm{re}\:\mathrm{typing}\:\mathrm{faster}\:\mathrm{than}\:\mathrm{me}\:;−\right) \\ $$

Commented by Rasheed.Sindhi last updated on 18/May/18

Only for this time.Always you beat me!

$$\mathrm{Only}\:\mathrm{for}\:\mathrm{this}\:\mathrm{time}.\mathrm{Always}\:\mathrm{you}\:\mathrm{beat}\:\mathrm{me}! \\ $$

Answered by Rasheed.Sindhi last updated on 18/May/18

Q_2 a=32 S_∞ =(a/(1−r))=48 ((32)/(1−r))=48 32=48−48r r=((48−32)/(48))=(1/3) t_n =ar^(n−1) t_8 =32×(1/3)^(8−1) =((32)/3^7 )=((32)/(2187))

$$\mathrm{Q}_{\mathrm{2}} \\ $$$$\:\:\:\:\:{a}=\mathrm{32} \\ $$$$\:\:\:\:{S}_{\infty} =\frac{{a}}{\mathrm{1}−{r}}=\mathrm{48} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{32}}{\mathrm{1}−{r}}=\mathrm{48} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{32}=\mathrm{48}−\mathrm{48}{r} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{r}=\frac{\mathrm{48}−\mathrm{32}}{\mathrm{48}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:{t}_{{n}} ={ar}^{{n}−\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:{t}_{\mathrm{8}} =\mathrm{32}×\left(\mathrm{1}/\mathrm{3}\right)^{\mathrm{8}−\mathrm{1}} =\frac{\mathrm{32}}{\mathrm{3}^{\mathrm{7}} }=\frac{\mathrm{32}}{\mathrm{2187}} \\ $$

Answered by MJS last updated on 18/May/18

(x−α)(x−β)=x^2 −(α+β)x+αβ we know that αβ=4 and α+β=3 ⇒ α^2 +β^2 =(α+β)^2 −2αβ=3^2 −2×4=1 (x−(1/α^2 ))(x−(1/β^2 ))=x^2 −((α^2 +β^2 )/(α^2 β^2 ))x+(1/(α^2 β^2 ))= =x^2 −(1/(16))x+(1/(16)) 16x^2 −x+1=0

$$\left({x}−\alpha\right)\left({x}−\beta\right)={x}^{\mathrm{2}} −\left(\alpha+\beta\right){x}+\alpha\beta \\ $$$$\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:\alpha\beta=\mathrm{4}\:\mathrm{and}\:\alpha+\beta=\mathrm{3}\:\Rightarrow \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta=\mathrm{3}^{\mathrm{2}} −\mathrm{2}×\mathrm{4}=\mathrm{1} \\ $$$$\left({x}−\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\right)\left({x}−\frac{\mathrm{1}}{\beta^{\mathrm{2}} }\right)={x}^{\mathrm{2}} −\frac{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }{\alpha^{\mathrm{2}} \beta^{\mathrm{2}} }{x}+\frac{\mathrm{1}}{\alpha^{\mathrm{2}} \beta^{\mathrm{2}} }= \\ $$$$={x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{16}}{x}+\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$\mathrm{16}{x}^{\mathrm{2}} −{x}+\mathrm{1}=\mathrm{0} \\ $$

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